Let $f:I \to X$ be a loop at $f(0)=f(1)=x_0$. Suppose that $f':S^1 \to X$ be a continuous map so that $f'\circ \exp =f$, where $\exp:I\to S^1$ be the exponential map.
My question is that:
If $f\simeq g$ (free homotopy), then $f' \simeq g'$ (free homotopy)?
where $f(0)=f(1)=g(0)=g(1)=x_0$.
Here is a counterexample. Let $X = S^1$ which is not contractible. Let $f' =id$ and $g'$ be the constant map $g'(x) = x_0$.Then $f'$ and $g'$ are not freely homotopic. However, the maps $f = f' \circ \exp$ and $g = g' \circ \exp$ are freely homotopic because their domain $I$ is contractible.
Note that any two maps $f, g : I \to X$ are freely homotopic.